P= F x v

[plzNOCarribbean]

Can someone please explain the meaning behind this equation to me. I understand that power is the rate of energy expenditure, or how much work is being done but I don't understand when to substitute this equation for the formal P=work/time.

I understand that work=F x d and d/t is equal to v, which is where the v (speed) comes from. An example that I tried to make up is as follows:


If we apply an INITIAL FORCE 20 N force to a box that's at rest, and then we remove that force, the power, or the amount of work that we did ON the box is equal to the initial force that we provided x the SPEED of the box, which must be CONSTANT.

why must the speed be constant? because if the speed of the box is not constant, than it must be accelerating, which it certainly cannot be since we removed our initial force. is this line of reasoning correct?

Do you use this equation when your not given the time interval with which a certain amount of work is done, but instead are given the force used to conduct that work and the velocity of the object as a result of the force?

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[MT Headed]

This equation works best when there is one force pushing on an object and there is another force pushing on it in the opposite direction with the same magnitude. Think of a car driving into the wind at max speed, or a feather that is falling at constant speed instead of continuing to accelerate.

The device doing the pushing (say your car engine) is applying a force, over a distance, in a specified amount of time. It delivers power. Sometimes it is easier to calculate power delivered as (force) x (distance / time) instead of the usual (force x distance) / (time).

You are hovering right at the edge of where a calculus-based physics answer would make more sense, but calculus-based physics is not required for the MCAT.

 

[plzNOCarribbean]

This equation works best when there is one force pushing on an object and there is another force pushing on it in the opposite direction with the same magnitude. Think of a car driving into the wind at max speed, or a feather that is falling at constant speed instead of continuing to accelerate.

The device doing the pushing (say your car engine) is applying a force, over a distance, in a specified amount of time. It delivers power. Sometimes it is easier to calculate power delivered as (force) x (distance / time) instead of the usual (force x distance) / (time).

You are hovering right at the edge of where a calculus-based physics answer would make more sense, but calculus-based physics is not required for the MCAT.

Yeah, not trying to hover around the calculus based physics. your explanation does make sense though. But more specifically, would you normally use this when you are not given a time interval in the problem but instead only given the force applied to an object and the speed with which the object travels as a result of that force?

 

[bullsfan1986]

Can someone please explain the meaning behind this equation to me. I understand that power is the rate of energy expenditure, or how much work is being done but I don't understand when to substitute this equation for the formal P=work/time.

I understand that work=F x d and d/t is equal to v, which is where the v (speed) comes from. An example that I tried to make up is as follows:


If we apply an INITIAL FORCE 20 N force to a box that's at rest, and then we remove that force, the power, or the amount of work that we did ON the box is equal to the initial force that we provided x the SPEED of the box, which must be CONSTANT.

why must the speed be constant? because if the speed of the box is not constant, than it must be accelerating, which it certainly cannot be since we removed our initial force. is this line of reasoning correct?

Do you use this equation when your not given the time interval with which a certain amount of work is done, but instead are given the force used to conduct that work and the velocity of the object as a result of the force?

Well, if you're saying that the total work done on the box=to the initial force applied on the box and F x d = W so for your example 20N x d= 20 Nxm where d= 1.

Now that you have distance you can solve for time by using a kinematics equation where:

Vf^2= Vi^2 + 2ax and because Vi=0...Vf^2= 2ax. Because you said that the problem gives you a velocity (Ill say that vf=4 for simplicity) you can solve for a and get 8m/s^2.

then you can use the equation and solve for t where v=vo+at where t= 0.5s.

Now that you know that your change in work is = 20Nxm..you can calculate for power where P= 20Nxm/0.5s....so power = 40Nxm/s

I hope I understood your question correctly and didn't mistakenly misread something from what you were trying to ask.

 

[MT Headed]

Yeah, not trying to hover around the calculus based physics. your explanation does make sense though. But more specifically, would you normally use this when you are not given a time interval in the problem but instead only given the force applied to an object and the speed with which the object travels as a result of that force?

Yes, assuming the speed is constant (like in my examples). This formula wouldn't apply if the speed was changing. I suppose it would work if the speed was changing and you plugged in the average speed, but again we are pushing the limits of what can be done in the algebra based physics on the MCAT.

 

[MD Odyssey]

Personally, I find the definition of power as force times velocity to be a gross simplification that is, in a lot of situations, misleading, because it implicitly requires that the velocity of the particle be constant. There are lots of physical situations where I might ask for the power applied to do a particular task in which the velocity is not constant. I can even cook up a simple example where the acceleration might not even be constant. That simplistic interpretation of power would be completely useless.

Consider the case of a proton being accelerated by an electric field in a linear particle accelerator. Let's further assume that the acceleration is uniform and that we know the final velocity and the length of the distance over which we applied the force. If we wish to find the power required, the idea of force times velocity as power is of no help to us. We don't know the strength of the electric field, so even if we had a constant velocity, which we don't, we still couldn't find the power. A clever test-writer might actually tell you the strength of the electric field or tell you what the force is that the field exerts on a proton and then lay a trap by inserting a distractor in the answer choices that multiplies that force by the final velocity. That would probably nail a lot of people that thought of power as force times velocity.

A more fundamental approach would be to rely on the definition of power as energy expended over time, as the OP stated. In this case, the work done on the particle is simply the change in its kinetic energy, which is easy to find, since we know the final velocity and the mass of the proton. Finding the power used at this point would be trivial.

Power also shows up in other places where forces aren't even active at a macroscopic level. As an example, consider the case of a resistive element like a wire or heating element - understanding power as it pertains to forces is nonsense there. But, seeing it as the amount of energy expended over a particular amount of time is totally clear. Other examples of this sort of thing are legion, but I'd imagine you get the idea.

Think of power as the amount of energy expended over a duration of time, and you will never go wrong.

 

[MD Odyssey]

Let me answer your question specifically too:

Can someone please explain the meaning behind this equation to me. I understand that power is the rate of energy expenditure, or how much work is being done but I don't understand when to substitute this equation for the formal P=work/time.

Power as a rate of energy expenditure is a far more accurate and complete way to look at power. Power as force times velocity is only true for specific situations. A lot of the confusion about work stems from the fact that it is actually defined in a more complicated way - a line integral of the force along the path of the displacement. Since most people don't know what that is, only certain types of problems are given in algebra-based physics courses. This is why physics seems so complicated to pre-meds that haven't had a calculus-based physics course - it seems like a deluge of equations which have to be memorized along with when one can or cannot use them. Fundamentally speaking, think of power as the energy expended divided by the time required, and you'll never go wrong on the MCAT.

 

[plzNOCarribbean]

Let me answer your question specifically too:



Power as a rate of energy expenditure is a far more accurate and complete way to look at power. Power as force times velocity is only true for specific situations. A lot of the confusion about work stems from the fact that it is actually defined in a more complicated way - a line integral of the force along the path of the displacement. Since most people don't know what that is, only certain types of problems are given in algebra-based physics courses. This is why physics seems so complicated to pre-meds that haven't had a calculus-based physics course - it seems like a deluge of equations which have to be memorized along with when one can or cannot use them. Fundamentally speaking, think of power as the energy expended divided by the time required, and you'll never go wrong on the MCAT.

yeah, I agree completely. I just wanted to see why all the prep book included this alternative equation in their texts, just incase a situation occurred where it had to be used. Alright, Il take your advice with regards to thinking of it as work/time, or the rate of energy expenditure.

And with your example regarding circuit elements expending energy,ie, the reason light bulbs heat up after they've been on for so long. I understand that force is useless here and that you have to rely on the electrostatic approach to figuring out the power being supplied by each circuit element ( P= IV, P= I^2R, V^2/R) and I remember specific examples that give the power and ask how much heat is dissipated over a certain given time, by which you multiply Pxt= W. Thanks for the clarification on why to use the initial equation.

 

[MT Headed]

I just wanted to see why all the prep book included this alternative equation in their texts, just incase a situation occurred where it had to be used.

I can say with certainty that this equation does show up on an AAMC practice test. A passage set up a specific, contrived situation much like the scenarios outlined above, and then a question asked something about power.

If you recognized that this is one of those rare P=Fv situations, the answer was obvious and fast. If you approached it from the more general P=energy/time you would waste a lot of time and you would likely get the wrong answer because the other answer choices involved energy and time, and were almost correct.

It's little nuggets like these that separate the 11's from the 12+.

 

[plzNOCarribbean]

I can say with certainty that this equation does show up on an AAMC practice test. A passage set up a specific, contrived situation much like the scenarios outlined above, and then a question asked something about power.

If you recognized that this is one of those rare P=Fv situations, the answer was obvious and fast. If you approached it from the more general P=energy/time you would waste a lot of time and you would likely get the wrong answer because the other answer choices involved energy and time, and were almost correct.

It's little nuggets like these that separate the 11's from the 12+.

Really? thanks for your help MT headed. So, would it apply then to a scenario when something is falling for instance, and so the force causing the object to fall would be mg, while another force , say air resistance, opposes the objects fall so that the velocity is constant? you said it applies only when one force is opposing, or balancing out another force so that v=constant. is that correct?

could you provide me with a few specific examples

 

[MT Headed]

I don't want to spoil any AAMC surprises, but you seem to have the hang of it. You could try looking in a algebra-based physics textbook where they introduce P=Fv and see what examples they come up with.

 

[MD Odyssey]

I can say with certainty that this equation does show up on an AAMC practice test. A passage set up a specific, contrived situation much like the scenarios outlined above, and then a question asked something about power.

If you recognized that this is one of those rare P=Fv situations, the answer was obvious and fast. If you approached it from the more general P=energy/time you would waste a lot of time and you would likely get the wrong answer because the other answer choices involved energy and time, and were almost correct.

It's little nuggets like these that separate the 11's from the 12+.

You're totally correct. I ran across a problem like this in one of the materials I was using. Calculating the work done and dividing by time didn't really take that much longer, but then, the physics sections tends to come a bit more naturally for me.

As long as you understand the limitations of power as force times velocity, then there's nothing wrong using that equation. Indeed, the more general expression reduces to precisely that. I just didn't want to confuse you.